Birth of Scientific ComputationsMartin Gander and Gerhard Wanner, Genève

L. Euler W. Ritz B.G. Galerkin

Brissago, 30 sept. 2011.– p.1/73

“c’est des calculs monstrueux. . . heureusement on a desordinateurs. . .”

(R. Durrer, giovedi, ora 9:44)

“ . . . heureusement on a des bonnes méthodes numériques. . .”

(addendum G. Wanner, venerdi, ora 10:34)

– p.2/73

Euler (E122,1747):Differential Eqs. for Mechanics.

“While physicists call these “Newton’s equations”,they occur nowhere in the work of Newtonor of anyone else prior to 1747.”“... such is the universal ignoranceof the true history of mechanics.”

(C. Truesdell,Essays in the History of Mechanics,1968)

– p.3/73

Euler (E122,1747):Differential Eqs. for Mechanics.

“While physicists call these “Newton’s equations”,they occur nowhere in the work of Newtonor of anyone else prior to 1747.”“... such is the universal ignoranceof the true history of mechanics.”

(C. Truesdell,Essays in the History of Mechanics,1968)

Easier:First order eq.

dy

dx= V (x, y) 1

−1

1

dydx = x2 + y2

x0, y0

Initial value

x0, y0

Solution

– p.3/73

“Euler’s Method” ( E342), Inst. Calc. Integralis 1768, §650:

dydx = V (x, y)

xn+1 = xn + h,

yn+1 = yn + hV (xn, yn)

(Preuve de convergence:Cauchy 1824)

1

−1

1

x0, y0

x1, y1

x2, y2

h=1/4 h=1/8

h=1/32

– p.4/73

Higher Order Method (Euler E342), ICI 1768, §656:

Differentiate for higher derivativesEx.: y′ = x2 + y2:

– p.5/73

1

−1

1

x0, y0

x1, y1

x2, y2

h = 1/4 h=1/8

h=1/32

dydx = x2 + y2, Euler

1

−1

1

x0, y0x1, y1

x2, y2

h=1/2

h=1/4

h=1/16

dydx = x2 + y2, Taylor 2

yn+1 = yn + hy′n yn+1 = yn + hy′n + h2

2y′′n

– p.6/73

1

−1

1

x0, y0x1, y1

x2, y2

h=1/2

h=1/4

h=1/16

dydx = x2 + y2, Taylor 2

1

−1

1

x0, y0

x1, y1

x2, y2

h=1/2

h=1/4

dydx = x2 + y2, Taylor 3

yn+1 = yn + hy′n + h2

2y′′n yn+1 = yn + hy′n + h2

2y′′n + h3

6y′′′n

– p.7/73

“Automatic differentiation”. (A. Gibbons 1960,E. Fehlberg 1964, R.E. Moore 1966, W. Gautschi 1966,. . . )Example:y′ = x2 + y2

Sety(x0+h) = y0 + hy1 + h2y2 + . . . , x2 = x20 + 2x0h+ h2,

develop: y′ = y1 +2y2h +3y3h2 +4y4h

3 + . . .

= x20 +2x0h +h2

+y20 +2y1y0h +2y0y2h

2 +2y0y3h3

y21h

2 +2y1y2h3 + . . .

⇒ y1 = x20 +y2

0, 2y2 = 2x0 +2y1y0, 3y3 = 1+2y0y2 +y21, . . .

– p.8/73

“Automatic differentiation”. (A. Gibbons 1960,E. Fehlberg 1964, R.E. Moore 1966, W. Gautschi 1966,. . . )Example:y′ = x2 + y2

Sety(x0+h) = y0 + hy1 + h2y2 + . . . , x2 = x20 + 2x0h+ h2,

develop: y′ = y1 +2y2h +3y3h2 +4y4h

3 + . . .

= x20 +2x0h +h2

+y20 +2y1y0h +2y0y2h

2 +2y0y3h3

y21h

2 +2y1y2h3 + . . .

⇒ y1 = x20 +y2

0, 2y2 = 2x0 +2y1y0, 3y3 = 1+2y0y2 +y21, . . .

Surprise:Euler invented this too !! (E342), §663.– p.8/73

Second order equations(EulerE366, 1769,§1082)

ddx

dt2= F (t, x) ⇒ dx

dt= v,

dv

dt= F (t, x) ⇒

tn+1 = tn + h, xn+1 = xn + hvn, vn+1 = vn + hF (tn, xn),

– p.9/73

Example: Particle attr. by two fixed centers (EulerE301, 1760)

ddx

dt2= − Ax

v3− B(x− a)

u3,

ddy

dt2= − Ay

v3− By

u3

v =√

x2 + y2, u =√

(x− a)2 + y2 A = 2, B = 1, a = 1

−1

1

A B

C init. positioninit. velocity

– p.10/73

Euler’s method:

xn+1 = xn + hvn, vn+1 = vn + hF (tn, xn),

−1

1

A B

C

h = 1/4

– p.11/73

decrease step size:

xn+1 = xn + hvn, vn+1 = vn + hF (tn, xn),

−1

1

A B

CCC

h = 1/32, 1/64, 1/128.

– p.12/73

Learn from Newton(look at Kepler problem):

S

hv0

hv1

h2F0

x0

x1

c

x2 S

AB

c

C

Eulerxn+1 = xn + hvn,

vn+1 = vn + hF (tn, xn).

(areas not preserved)

Newton“symplecticEuler” methodxn+1 = xn + hvn,

vn+1 = vn + hF (tn+1, xn+1).

(areas preserved)(is geometric integrator).

Funny: “Newton’s equations” are due to Euler...“Symplectic Euler method”is due to Newton. – p.13/73

Symplectic Euler method at two fixed centers:

−1

1

A B

CCC

h = 1/128

– p.14/73

Third Order Method (Euler E342), ICI 1768, §656:

−1

1

A B

CCC

h = 1/32, 1/64, 1/128

– p.15/73

What is Symplecticity? (Poincaré 1899)vary initial values⇒ preserves (certain sums of) areas.Example: Symplectic Euler method:

ϕt

SE

p = −hHq(p, q)

q = hHp(p, q)⇒

pn+1 = pn − hHq(pn+1, qn)

qn+1 = qn + hHp(pn+1, qn)– p.16/73

The Stormer-Verlet Method.Störmer 1907 (Astronomy), Verlet 1967 (Molecular dynamics)

ϕt

SE

SE*

• Strang splitting :

• Composition of SE and SE∗⇒ symplectic.

– p.17/73

Example: Outer solar system.

J SUN

P

explicit Euler,h = 10

J SUN

P

implicit Euler,h = 10

J SUN

P

symplectic Euler,h = 100

J SUN

P

Störmer–Verlet,h = 200

Pour plus de détails écoutezconférences deJacques Laskar...

ou lisez .......– p.18/73

II. Ritz-Galerkin method:

Problem. Appartment inCanadian winter at−20

0 1 2 3 4 5 6 7 8 9 10 11 120

1

2

3

4

5−20 −20 −20 −20

(Martin Gander)

Find temperature everywhere !!

– p.19/73

Solution: Fourier’s Heat Equation (1807, 1822)

vi−1 vi vi+1

dvi

dt= K · ((vi+1 − vi)− (vi − vi−1))

Heat equation for 1-dimensional boby. – p.20/73

In two dimensions:

h hh

hu0 u1u2

u3

u4

∂u0

∂t=u1 + u2 + u3 + u4 − 4u0

h2+ f0.

continuous problemh→ 0:

∂u

∂t=∂2u

∂x2+∂2u

∂y2+ f(x, y) equilib.:

−∆u = f

(f = source of heat).

How to solve ???

– p.21/73

In two dimensions:

h hh

hu0 u1u2

u3

u4

∂u0

∂t=u1 + u2 + u3 + u4 − 4u0

h2+ f0.

continuous problemh→ 0:

∂u

∂t=∂2u

∂x2+∂2u

∂y2+ f(x, y) equilib.:

−∆u = f

(f = source of heat).

How to solve ???“Remember, Rome was not built in a day !!”(Laurel & Hardy,Dirty Work 1933)

– p.21/73

1700

1800

1900

GalileiGalileiBernoulliBernoulli

EulerEulerLagrangeLagrange

Euler ICIEuler ICILaplaceLaplace

DirichletDirichlet ThomsonThomsonGaussGaussRiemannRiemann

WeierstrassWeierstrass SchwarzSchwarz

HilbertHilbertRitzRitz TimoshenkoTimoshenko BubnovBubnov

GalerkinGalerkinCourantCourant

LFLZTC. . .LFLZTC. . .

GermaineGermainePoissonPoisson

KirchhoffKirchhoff

RayleighRayleigh

France,Italy Switz., Germany Russia – p.22/73

Galilei. Discorsi 1638, Giornata terza, Teorema 22 e Scolie:

“Da quanto si è dimostrato sembra si possa ricavare che ilmovimento più veloce da estremo ad estremo non avviene lungola linea più breve, cioè la retta, ma lungo un arco di cerchio.”

AB

C

D

E

F

G

“... Mr. Leibnits remarque en Galiléedeux fautes considerables: c’est quecet homme-là, qui étoit, sans contredit,le plus clairvoyant de son temsdans cette matiére, vouloit conjecturerque la courbe de la chainetteétoit une Parabole, que celle de laplus vite descente étoit un Cercle.."(Joh. Bernoulli, 1697)

– p.23/73

Bernoulli’s Brachystochrone.PROBLEMA NOVUM Ad cujus solutionem Mathematiciinvitantur. Datis in plano verticali duobus punctisA & B,assignare MobiliM viamAMB, per quam gravitate suadescendens, & moveri incipiens a punctoA, brevissimo temporeperveniat ad alterum punctumB.

(Joh. Bernoulli, Acta Erud., Jun. 1696)

dxdx

dydydsds

A

BM

x

y

∫

√

dx2 + dy2

√y

= min.

∫

√

1 + p2

ydx = min. (p = y′).

“on fait tout avec les actions. . . en physique. . . tout !”

(K. Zuleta, giovedi, ore 11:32)– p.24/73

Euler (E65, 1744):

J =

∫ b

a

Z dx = min! vel max! whereZ = Z(x, y, p), p =dy

dx

Euler’s Solution.

1. Approximate curveby polygon

2. Approx. Integralby -sum(“Riemann”)and diff. w.r. toν;

3. Set derivative to zero;

4. inverseEuler method⇒ (N = ∂Z∂y ,

P = ∂Z∂p ).

– p.25/73

Example 1.

J =

∫ b

a

(

p2

2− f · y

)

dx −→ min

−d2y

dx2= f(x) “Newton’s eqs.”

−1 1

A By(x)

f(x)

Example 2. (the Brachystochrone ; EulerE65, Caput II, §34).√

1 + p2

√2gy

− p2

√2gy

√

1 + p2= C or 1 =

√

1 + p2√

2gy · C .

same integral as in M. Kunz’ talk on expansion of the Univers⇒ Cycloide.

– p.26/73

Variational Calc. 1744E65 Inst. Calc. Integralis 1768E342– p.27/73

Euler’s “World”:

Variational Problems to Solve

Euler’s Differential Equations to Solve

Analytic Formulasor Numerical Methods for Solution

EulerE65 (1744)

EulerE342 (1768)

– p.28/73

Joseph Louis de Lagrange1755:

Ludovico de la Grange Tournier(19 years) writes 12 aug. 1755to Vir amplissime atque celeberrime L. Euler. From whomVir praestantissime atque excellentissime Lagrangereceives a kind and enthusiastic answer (6 september 1755).

Lagrange simplifies the derivation of Euler’s equations by theVariational Calculusnotationδy, whichvaries all values ofysimultaneously:

(Picture from

Euler, ICI

vol. 3, 1770,

Appendix)

A

B

x

y

v

ba

EulerE420(1772)– p.29/73

Euler E420(1772)“Methodus nova et facilis. . .”

J(ε) =

∫ b

a

Z(x, y + εv, p+ εv′) dx = min!|ε=0 Variat. Probl.

Differentiate:∂J(ε)

∂ε|ε=0 =

∫ b

a

(N · v + P · v′) dx = 0.

Integration by parts:∫ b

a

(N − d

dxP ) · v · dx = 0 “weak solution”

if v(a) = v(b) = 0. Hence, sincev is arbitrary,⇒

N − d

dxP = 0 “strong solution”.

– p.30/73

1700

1800

1900

BernoulliBernoulli

EulerEulerLagrangeLagrange

Euler ICIEuler ICILaplaceLaplace

DirichletDirichlet ThomsonThomsonGaussGaussRiemannRiemann

WeierstrassWeierstrass SchwarzSchwarz

HilbertHilbertRitzRitz TimoshenkoTimoshenko BubnovBubnov

GalerkinGalerkinCourantCourant

LFLZTC. . .LFLZTC. . .

GermaineGermainePoissonPoisson

KirchhoffKirchhoff

RayleighRayleigh

France, Italy Switz., Germany Russia – p.31/73

“Dirichlet’s Principle” (C.F. Gauss, Werke 5, p. 195, 1839;W. Thomson, Liouville J. 12 (1847) p. 496).∫∫

Ω(1

2((∂u

∂x)2 + (∂u∂y )2)− f · u) dx dy = min! ⇒ −∆u = f.

Proof in Eulerian style: Derive

h hh

hu0 u1u2

u3

u4

1

2

(

...+(u1−u0

h)2+(

u0−u2

h)2+(

u3−u0

h)2+(

u0−u4

h)2+...

)

−f0·u0...

with respect tou0 and set derivative= 0:

−u1 − u2 − u3 − u4 + 4u0

h2− f0 = 0. QED.

– p.32/73

B. Riemann (Thesis 1851) f(z) = u(x, y) + iv(x, y)

⇒ f ′ =

(

A −BB A

)

⇒ complex mapping is conformal (angle preserving).“ ... und ihre entsprechenden kleinsten Theile ähnlich sind;” (Thesis §21)

zz

H

α1z1

α2α3

α4z4

α5z5

α6

α7

α8

w = f(z)w = f(z)

f(H)

α1w1

α2

α3

α4

w4

α5w5

α6

α7

α8

– p.33/73

Riemann mapping theorem: (Thesis §21)

“Zwei gegebene einfach zusammenhängende

Flächen können stets so aufeinander bezogen

werden, dass jedem Punkte der einen Ein mitihm stetig fortrückender Punkt entspricht...;”

⇔

(drawing M. Gutknecht 18.12.1975)– p.34/73

Sketch of Proof: (Thesis §21; Th. Abelscher Funktionen 1857)

1. Placelog(z − z0);2. correct Relog |z − z0| − u(x, y),u harmonic, boundary val.→ 0

←−f

3. complete−iv(x, y) to holom. fcn.4. f = exp (log(z − z0)− u− iv)).

– p.35/73

Riemann’s Challenge:

Find harmonic functions∆u = 0 on any domainΩwith prescribed boundary conditionsu = F for (x, y) ∈ ∂Ω ?

xxxx

yyyy

w(x, y)

easy for rectangle and circle (Fourier, Poisson), but difficult forarbitrary domains...

– p.36/73

Riemann’s Audacious “Proof”:“Hierzu kann in vielen Fällen. . . ein Princip dienen, welches

Dirichlet zur Lösung dieser Aufgabe für eine der Laplace’schen

Differentialgleichung genügende Function. . . in seinen

Vorlesungen. . . seit einer Reihe von Jahren zu geben

pflegt.” (RiemannCrelle J.1857,Werkep. 97)

Idea.For all functions defined on a given domainΩ with theprescribed boundary values, the integral

J(u) =

∫∫

Ω

1

2

(

u2x + u2

y

)

dx dy is always> 0.

Choose among these functions the one for which this integralisminimal !(see citation; from here originates the name“DirichletPrinciple”and“Dirichlet boundary conditions”).

– p.37/73

Euler’s “World” Upside Down !!

Variational Problem∫

(u2x + u2

y)dxdy = min!

Differential Equation ∆u = 0

Dirichlet’s Principle

– p.38/73

Weierstrass’ Critics: (1869, Werke 2, p. 49)∫ 1

−1

(x · y′)2 dx = min!

y(−1) = a, y(1) = b

y = a+b2

+ b−a2

arctanx

ǫ

arctan 1

ǫ

−1 0 1a

b

“Die Dirichlet’sche Schlussweise führt also in dem betrachtetenFalle offenbar zu einem falschen Resultat.”

Riemann’s Answer to Weierstrass:“... meine Existenztheoreme sind trotzdem richtig”.

(see F. Klein,Entw. Math. 19. Jahrh., p. 264).

and Helmholtz:“Für uns Physiker bleibt das DirichletschePrinzip ein Beweis”.

... but not for most mathematicians ...

– p.39/73

Existence Proof without Dirichlet Principle:

H.A. Schwarz(1870, Crelle 74, 1872)Solve alternatively inΩ1 andΩ2

new boundary values on dotted curves⇒ convergesadd third, fourth domain etc.“Alternating method”

Ω1Ω2

– p.40/73

Rehabilitation of Dirichlet Principle:C. Arzelà(“primo tentativo” 1897);D. Hilbert (“trionfare” 1901, Annalen 1904, Crelle J. 1905):

“... eine besondere Kraftleistung beweisender Mathematik, ...”(F. Klein, Entw. Math. 19. Jahrh., p. 266).

Hilbert: “Das Dirichletsche Prinzip verdankte seinen Ruhm deranziehenden Einfachheit seiner mathematischen Grundidee,dem unleugbaren Reichtum der möglichen Anwendungen ...und der ihm innewohnenden Überzeugungskraft.”

“Mittlerweile war das verachtete und scheintoteDirichletschePrinzip durch Hilbert wieder zum Leben erweckt worden;...”(Hurwitz-CourantFunktionentheorie, Springer Grundlehren 3, p. 392)

⇒ Levi, Fubini, Lebesgue, Zaremba, Tonelli, Courant . . .

– p.41/73

1700

1800

1900

BernoulliBernoulli

EulerEulerLagrangeLagrange

Euler ICIEuler ICILaplaceLaplace

DirichletDirichlet ThomsonThomsonGaussGaussRiemannRiemann

WeierstrassWeierstrass SchwarzSchwarz

HilbertHilbertRitzRitz TimoshenkoTimoshenko BubnovBubnov

GalerkinGalerkinCourantCourant

LFLZTC. . .LFLZTC. . .

GermaineGermainePoissonPoisson

KirchhoffKirchhoff

RayleighRayleigh

France, Italy Switz., Germany Russia – p.42/73

Another Problem: The Elastic Plate.

(S. Germaine 1811/13/15, corr. J.L. Lagrange, S.D. Poisson1829, Kirchhoff (Crelle J. 40, 1850, p. 51-88))Problem from the Academy of Paris for 1907:

Ritz, had worked with many of such problems in his thesistrying to explain theBalmer seriesin spectroscopy (1902).– p.43/73

Ritz’s method ...

... transforms the problem into avariational problem(“wie man ohne weiteres einsieht”) ...

⇓

“Es ist ...J die potentielle Energie (... derKirchhoffscheAusdruck...)” – p.44/73

Ritz’s method ...... and approaches the solutiongloballyby a linear combinationof well chosen basis functionsψ1, ψ2, ψ3, . . .

⇓

⇓

... transforming the problem into afinite dimensionalminimization problem.

– p.45/73

Explicitly:

aij =

∫ 1

−1

∫ 1

−1

∆ψi ·∆ψj dx dy , bi =

∫ 1

−1

∫ 1

−1

ψi · f dx dy .

Jm =1

2

m∑

i,j=1

aijαiαj −m

∑

i=1

biαi ⇒ min ifm

∑

j=1

aijαj = bi

– p.46/73

First choice of basis functions(R = square−1, . . . 1):

ψ1(x, y) = (1− x2)2(1− y2)2

ψ2(x, y) = (1− x2)2(1− y2)2(x2 + y2)

ψ3(x, y) = (1− x2)2(1− y2)2(x4 + y4)

ψ4(x, y) = (1− x2)2(1− y2)2x2y2

ψ5(x, y) = (1− x2)2(1− y2)2(x6 + y6)

ψ6(x, y) = (1− x2)2(1− y2)2(x4y2 + x2y4) . . .

... leading to the system

53.4988 9.7271 2.1339 0.5404 0.6374 0.3048 = 1.1378

9.7271 21.9482 10.9187 1.7459 5.9936 1.2597 = 0.3251

2.1339 10.9187 9.6331 0.9095 7.1004 1.0177 = 0.1084

0.5404 1.7459 0.9095 0.4724 0.5147 0.3955 = 0.0232

0.6374 5.9936 7.1004 0.5147 6.3226 0.7312 = 0.0493

0.3048 1.2597 1.0177 0.3955 0.7312 0.4425 = 0.0155– p.47/73

53.4988 9.7271 2.1339 0.5404 0.6374 0.3048 = 1.1378

9.7271 21.9482 10.9187 1.7459 5.9936 1.2597 = 0.3251

2.1339 10.9187 9.6331 0.9095 7.1004 1.0177 = 0.1084

0.5404 1.7459 0.9095 0.4724 0.5147 0.3955 = 0.0232

0.6374 5.9936 7.1004 0.5147 6.3226 0.7312 = 0.0493

0.3048 1.2597 1.0177 0.3955 0.7312 0.4425 = 0.0155

“Cramer’s Rule”(Maclaurin 1748, Cramer 1750)

would require the computation of5040 such terms.– p.48/73

However, because system isdiagonally dominant, already

53.4988α1 = 1.1378 ⇒ α1 =1.1378

53.4988= 0.02127

gives an acceptable solution:

xxxx

yyyy

u1(x, y)

m = 1, errmax= 0.04

– p.49/73

For higher precision Ritz computes iteratively

53.4988 = 1.1378

9.7271 21.9482 = 0.3251

2.1339 10.9187 9.6331 = 0.1084

0.5404 1.7459 0.9095 0.4724 = 0.0232

0.6374 5.9936 7.1004 0.5147 6.3226 = 0.0493

0.3048 1.2597 1.0177 0.3955 0.7312 0.4425 = 0.0155

and repeat ...

53.4988 9.7271 2.1339 0.5404 0.6374 0.3048 = 1.1378

9.7271 21.9482 10.9187 1.7459 5.9936 1.2597 = 0.3251

2.1339 10.9187 9.6331 0.9095 7.1004 1.0177 = 0.1084

0.5404 1.7459 0.9095 0.4724 0.5147 0.3955 = 0.0232

0.6374 5.9936 7.1004 0.5147 6.3226 0.7312 = 0.0493

0.3048 1.2597 1.0177 0.3955 0.7312 0.4425 = 0.0155– p.50/73

“wozu der Rechenschieber angewandt werden kann...Eine direkte Lösung durch Determinanten würde 5stelligeLogarithmentafeln erfordern”.

α1 = 0.020247 α2 = 0.005209 α3 = 0.000284

α4 = 0.006119 α5 = −0.000019 α6 = 0.000116

xxxx

yyyy

u1(x, y)

xxxx

yyyy

u2(x, y)

xxxx

yyyy

u3(x, y)

xxxx

yyyy

u4(x, y)

m = 1, errmax= 0.04 m = 2, errmax= 0.003

m = 4, errmax= 0.000017 m = 6 – p.51/73

... beautiful in theory —but not in practice...

∆ψ6 = 8x2(

1− y2)2 (

x4y2 + x2y4)

−8(

1− x2) (

1− y2)2 (

4x3y2 + 2xy4)

x

−4(

1− x2) (

1− y2)2 (

x4y2 + x2y4)

+(

1− x2)2 (

1− y2)2 (

12x2y2 + 2 y4)

+8(

1− x2)2y2

(

x4y2 + x2y4)

−8(

1− x2)2 (

1− y2) (

2x4y + 4x2y3)

y

−4(

1− x2)2 (

1− y2) (

x4y2 + x2y4)

+(

1− x2)2 (

1− y2)2 (

2x4 + 12x2y2)

must compute∫ 1

−1

∫ 1

−1∆ψ6 ·∆ψ6 dx dy = 3052404736

6898776885from this ...

and 20 others ...

– p.52/73

Dirichlet’s Principle.

∆u = 0 u|∂R = F

or, after a subtraction,

−∆u = f u|∂R = 0 .

We insert

um = α1ψ1(x, y) + . . .+ αmψm(x, y)

into

J =

∫∫

Ω

(1

2((∂u

∂x)2 + (

∂u

∂y)2)− f · u) dx dy = min!

modify the basis functions according to the new boundarycondition ...

– p.53/73

... with solution

xxxx

yyyy

u1(x, y)

xxxx

yyyy

u2(x, y)

xxxx

yyyy

u3(x, y)

xxxx

yyyy

u4(x, y)

m = 1, errmax= 0.05 m = 2, errmax= 0.01

m = 4, errmax= 0.0024 m = 6

– p.54/73

Ritz’ Second Great Paper:Ernst Florens Friedrich Chladni : Leipzig 1787.

– p.55/73

Walther Ritz (1909): Theorie der Transversalschwingungeneiner quadratischen Platte mit freien Rändern

“Die Differentialgleichungen und Randbedingungen für dietransversalen Schwingungen ebener, elastischer Platten mitfreien Rändern sind bekanntlich zuerst in teilweise unrichtigerForm von Sophie Germain und Poisson, in definitiver Gestaltaber von Kirchhoff im Jahre 1850 gegeben worden.”

Chladni figures correspond to eigenpairs of the bi-harmonicoperator

∆2w = λw in Ω := (−1, 1)2

∂∂x

(

∂2w∂x2 + (2− µ)∂2w

∂y2

)

= 0, ∂2w∂x2 + µ∂2w

∂y2 = 0, x = −1, 1∂∂x

(

∂2w∂y2 + (2− µ)∂2w

∂x2

)

= 0, ∂2w∂y2 + µ∂2w

∂x2 = 0, y = −1, 1

Here,µ is the elasticity constant.

– p.56/73

Ritz’ method now leads to an eigenvalue problem:

– p.57/73

2) How to solve the eigenvalue problemK~a = λ~a?

• setzen wirA0 = 1, und in erster Annäherungλ0 = 13.95.Dann ergeben die fünf letzten Gleichungen die übrigenAi.• Wir berechnen für dieAi eine erste Approximation, indemwir alle Glieder rechts vernachlässigen neben denDiagonalgliedern . . .• Ein oder zwei sukzessive Korrektionen genügen meist, umdie vierte Stelle bis auf wenige Einheiten festzustellen.

iterative linear algebra, iterative eigenvalue calculation ...Ritz is very modern in 1909 !!

– p.58/73

Chladni Figures Computed by Ritz

– p.59/73

– p.60/73

Frequency Table Computed by Ritz

– p.61/73

1700

1800

1900

BernoulliBernoulli

EulerEulerLagrangeLagrange

Euler ICIEuler ICILaplaceLaplace

DirichletDirichlet ThomsonThomsonGaussGaussRiemannRiemann

WeierstrassWeierstrass SchwarzSchwarz

HilbertHilbertRitzRitz TimoshenkoTimoshenko BubnovBubnov

GalerkinGalerkinCourantCourant

LFLZTC. . .LFLZTC. . .

GermaineGermainePoissonPoisson

KirchhoffKirchhoff

RayleighRayleigh

France, Italy Switz., Germany Russia – p.62/73

Ritz’ ideas⇒ Russia: S.P. Timoshenko (Kiev 1910):

Nous ne nous arrêterons plus sur le côté mathématique de cette

question: un ouvrage remarquable du savant suisse, M. Walter Ritz, . . .

– p.63/73

S.P. Timoshenko (1910):

– p.64/73

Ivan Bubnov (1872-1919)

Structural Mechanics of Shipbuilding

[Part concerning the theory of shells]

– p.65/73

Boris Grigoryevich Galerkin (1871-1945)

Beams and Plates

Series solution of some problems inelastic equilibrium of rods and plates

(Petrograd, 1915)

(Boris Grigoryevich Galerkin and the famous paper which is now quoted in the

literature for the invention of the “Galerkin” method)

– p.66/73

B.G. Galerkin (1915):

– p.67/73

Richard Courant(address to the AMS, on May 3rd, 1941)

“At first, the theoretical interest in existence proofs dominated, and onlymuch later were practical applications envisaged...

– p.68/73

First mention of Finite Element Method:

(Footnote in the first edition of the book by Hurwitz and Courant (1922) withfirst

sketch of triangular FE methodfor proof ofRiemann Mapping Theorem; removed

in second edition)– p.69/73

the base functionsψi(x, y) (compact support):

phii

phij

phik

x

u

y

and the linear combinationsα1ψ1 + α2ψ2 + α3ψ3 + . . .

⇒ matrixA has many zeros and is easy to calculate.– p.70/73

Difficulty: For large problems convergence of iterationsfor liner systemextremely slow.

Remedy: inspired by

H.A. Schwarz(1870, Crelle 74, 1872)Solve alternatively inΩ1 andΩ2

“Alternating method”

Ω1Ω2

⇒ Domain Decomposition Method

– p.71/73

... voila ...

∆u = f Ritz-Galerkin FE method Domain decomp.– p.72/73

Grazie !

– p.73/73